Question

Three unbiased coins are tossed together. Find the probability of getting
$\left(a\right)\left(i\right)\text{one tail}\left(ii\right)\text{two tails}\left(iii\right)\text{all tails}\left(iv\right)\text{at least two tails}$
$\left(b\right)\left(i\right)\text{at most two tails}\left(ii\right)\text{At most two heads}$

Answer 1

Here S=( TTT,TTH,THT,HTT,THH,HTH,HHT,HHH )
a(i) one tail (THH,HTH,HHH)
Then E=3
So (p)=$\frac{n\left(E\right)}{n\left(S\right)}=\frac{3}{8}$
a(ii) two tails
Then E=3
So (p)=$\frac{n\left(E\right)}{n\left(S\right)}=\frac{3}{8}$
a(iii) all tails
Then E=1
So (p)=$\frac{n\left(E\right)}{n\left(S\right)}=\frac{1}{8}$
a(iv) at least two tails
Then E=4
So (p)=$\frac{n\left(E\right)}{n\left(S\right)}=\frac{4}{8}=\frac{1}{2}$
(b) (i) at most two tails
Then E=7
So (p)=$\frac{n\left(E\right)}{n\left(S\right)}=\frac{7}{8}$
b(ii) At most two heads
Then E=7
So (p)=$\frac{n\left(E\right)}{n\left(S\right)}=\frac{7}{8}$